Question

If in \((0,2\pi)\), \(2\cos x+k=3\sec x\) and \(\cot x=-1\), then the sum of squares of all possible values of \(k\) is

• A. \(0\)
• B. \(4\sqrt2\)
• C. \(8\)
• D. \(16\)

Hint:

First use the trigonometric condition to determine all possible values of \(x\), then substitute each value into the given equation to obtain the corresponding values of the parameter.

Answer 1

The Correct Option is D

Step 1: Find the possible values of \(x\). Given \[ \cot x=-1, \] we have \[ x=\frac{3\pi}{4},\; \frac{7\pi}{4} \] in the interval \[ (0,2\pi). \] Their cosine values are \[ \cos\frac{3\pi}{4} =-\frac{1}{\sqrt2}, \qquad \cos\frac{7\pi}{4} =\frac{1}{\sqrt2}. \]

Step 2: Find the corresponding values of \(k\). From \[ 2\cos x+k=3\sec x, \] we get \[ k=3\sec x-2\cos x. \] For \[ x=\frac{3\pi}{4}, \] \[ k =3(-\sqrt2)-2\left(-\frac1{\sqrt2}\right) =-2\sqrt2. \] For \[ x=\frac{7\pi}{4}, \] \[ k =3(\sqrt2)-2\left(\frac1{\sqrt2}\right) =2\sqrt2. \]

Step 3: Find the required sum. Hence, \[ (-2\sqrt2)^2+(2\sqrt2)^2 =8+8 =16. \] Therefore, \[ \boxed{16}. \] Thus, \[ \boxed{(D)} \] is the correct answer.